07-Str-A1 · May 2018

Question 5 of 8: Influence Lines

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 07-Str-A1 Elementary Structural Analysis, National Examinations, May 2018. Three hours, CLOSED BOOK (an approved Casio or Sharp calculator is permitted). Six questions constitute a complete paper: Questions 1 to 5 are compulsory and one of Questions 6, 7 or 8 is chosen. Marks are shown against each question and total 100. Every question is worked here, including all three of the optional Questions 6, 7 and 8, so that the set can be used as a study resource.

Reference texts. R. C. Hibbeler, Structural Analysis (Ch. 2 stability and determinacy; Ch. 4 influence lines; Ch. 6 deflections by virtual work; Ch. 11 displacement method; Ch. 12 moment distribution); A. Kassimali, Structural Analysis (Ch. 3, 8, 13, 16); K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis; J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods. For Canadian design practice the companion documents are the National Building Code of Canada (Part 4, load combinations) and CSA S16 Design of Steel Structures; this paper is pure analysis, so no design code is invoked in the answers.

Check — readings taken from the printed figures. Every dimension, support symbol and load extent below was read from the printed figures. Four readings carry the answers and are stated explicitly so that a marker can check them: (i) in Question 2(a) the 10 kN/m block covers only the 2 m overhang, ending at the roller; (ii) in Question 4(a) the truss is a Warren truss whose top nodes are offset 3 m from the bottom nodes (the 3 m dimension at the top right confirms this); (iii) in Question 7 the 10.8 kN/m ruling starts directly above joint 1 and runs to joint 4, so it acts over the whole horizontal projection, inclined member included; (iv) in Question 8 the thin vertical line between U₂ and L₃ touches neither joint circle and is the witness line of the 3 m dimension, not a member — taking it as a member would make the truss statically indeterminate and the question unanswerable by virtual work.

Question 5: Influence Lines (20 marks: 9 + 11)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Influence lines for the frame

Given. A horizontal member A–D with a free end at A, a rigid junction at B where a 2 m column runs down to a pin at E, an internal hinge at C and a pin at D. The spacings are A–B = 1 m, B–C = 2 m and C–D = 6 m. A unit downward load travels along A–D.

Find. Influence lines for the bending moment and the shear force immediately to the right of B and for the horizontal reaction at E, with the largest absolute ordinate of each.

Question 5(a): the frame. Four reaction components (pins at D and E) against three equilibrium equations plus one hinge condition at C make it determinate.

Approach. Place the unit load at a general position $x$ measured from B, solve the four reaction components from the three equations of global equilibrium plus the hinge condition at C, and then read the required quantity from the free body on one side of the section; each influence line is piecewise linear, so only the values at A, B, C and D are needed.

Reaction at D from the hinge condition. The piece to the right of C is held by the pin at D alone, so taking moments about C for that piece, $$D_y = \frac{x - 2}{6}\ \ (2 \le x \le 8), \qquad D_y = 0\ \ (x < 2),$$ with $x$ measured from B. Vertical equilibrium then gives $E_y = 1 - D_y$.
Horizontal reactions. Moments about E for the whole frame, with D at $(8, 0)$ and E at $(0, -2)$, give $$8 D_y - 2 D_x - x = 0 \;\Longrightarrow\; D_x = \frac{8 D_y - x}{2}, \qquad E_x = -D_x.$$ Substituting the two ranges, $$E_x = \frac{x}{2}\ \ (x \le 2), \qquad E_x = \frac{8 - x}{6}\ \ (2 \le x \le 8).$$ The influence line for $E_x$ therefore runs from $-0.5$ at A through zero at B to a peak of $\boxed{+1.00}$ at the hinge C, falling linearly to zero at D.
Moment immediately right of B. Taking the free body to the left of the section — the overhang A–B, the column and the pin at E — and summing moments about B, $$M_{B^+} = x - 2E_x \ \ (x \le 0), \qquad M_{B^+} = -2E_x \ \ (x \ge 0).$$ On the overhang the two terms cancel identically, so the ordinate is zero for the whole of A–B; between B and C it falls linearly to $$M_{B^+}(C) = -2(1.00) = \boxed{-2.00\ \text{m}},$$ the largest absolute ordinate, and it returns linearly to zero at D. The influence line is entirely hogging.
Shear immediately right of B. The shear is the sum of the upward forces to the left of the section, and the column reaction $E_y$ is always on that side: $$V_{B^+} = E_y - 1 \ \ (x \le 0), \qquad V_{B^+} = E_y \ \ (x > 0).$$ On the overhang $E_y = 1$ and the two terms cancel, so the ordinate is again zero; the line jumps by the customary unit step at the section to $\boxed{+1.00}$, holds that value all the way to the hinge because $D_y$ is still zero there, and then falls linearly to zero at D.
Read the peaks. The greatest absolute ordinates are 2.00 m for the moment and 1.00 for both the shear and the horizontal reaction, all occurring with the unit load standing at the hinge C — which is the expected result, since a load at the hinge is carried entirely by the left-hand portion and therefore works the column hardest.

Question 5(a): the three influence lines. Each is piecewise linear with vertices at B, C and D; all three vanish for a unit load anywhere on the overhang A–B.

(b) Influence line for a truss diagonal and the governing vehicle position

Given. A truss of 24 m span in four 6 m panels and 5 m deep, pinned at L₁ and on a roller at L₅, with top joints U₁, U₂ and U₃ above L₂, L₃ and L₄. The long diagonal of interest, L₁–U₂, runs 12 m across and 5 m up, so its length is 13 m. The vehicle is three downward loads — 100 kN, then 100 kN at 2 m, then 50 kN a further 6 m behind — travelling from left to right along the bottom chord.

Find. The ordinates of the influence line for the force in L₁–U₂, and the maximum compression and maximum tension the vehicle can produce in that member.

Question 5(b): the truss. The diagonal L₁–U₂ crosses the vertical U₁–L₂ without connecting to it.

Approach. A vertical section between L₂ and L₃ cuts the top chord, the diagonal and the bottom chord; the two chords are horizontal, so vertical resolution isolates the diagonal and expresses its force as a multiple of the panel shear. Evaluating that at each panel point gives the ordinates, after which the vehicle is stepped through the breakpoints.

Express the diagonal force in terms of the panel shear. The section cuts U₁–U₂, L₁–U₂ and L₂–L₃; only the diagonal has a vertical component, of magnitude $5/13$ of its force. Vertical equilibrium of the left portion gives $$F = -\frac{13}{5}\left(V_{L1} - \textstyle\sum P_{\text{left}}\right) = -2.60 \times (\text{panel shear}),$$ with tension taken positive.
Evaluate at each panel point. A unit load standing on a support gives zero in every member, so the ordinates at L₁ and L₅ are both zero. For the load at L₂ the reaction is $0.75$ and the load itself lies to the left of the section, so the panel shear is $-0.25$ and $$y_{L2} = -2.60(-0.25) = +0.650 \ (\text{tension}).$$ For the load at L₃ the reaction is $0.50$ with nothing to the left of the section, giving $y_{L3} = -2.60(0.50) = -1.300$ (compression), and for the load at L₄ the reaction is $0.25$, giving $y_{L4} = -0.650$ (compression). Because the ordinate at L₄ is exactly half that at L₃, the influence line runs as a single straight line from the peak at L₃ to zero at L₅, which is precisely the shape printed on the examination paper.
Maximum compression. The largest compressive ordinate is $-1.300$ at L₃, so the leading heavy axle is placed there; the second 100 kN then stands at $x = 14$ m and the 50 kN at $x = 20$ m, where the ordinates interpolate to $-1.08333$ and $-0.43333$: $$F_c = 100(1.300) + 100(1.08333) + 50(0.43333) = 130.0 + 108.3 + 21.7 = \boxed{260.0\ \text{kN compression}}.$$ Trying either of the other two axles on the peak gives 227.5 kN and a tensile value, so this position governs.
Maximum tension. The only tensile region is L₁ to the zero at $x = 8$ m, so the vehicle must straddle it with the light axle beyond. Because the effect is piecewise linear in the vehicle position, the extreme occurs with an axle on a vertex: placing the leading 100 kN at $x = 4$ m puts the second 100 kN exactly on the tensile peak at L₂ and the 50 kN on the compressive peak at L₃, $$F_t = 100(0.43333) + 100(0.650) - 50(1.300) = 43.3 + 65.0 - 65.0 = \boxed{43.3\ \text{kN tension}}.$$ A search over every vertex position confirms that no other placement is more tensile.

[Figure not reproduced: Question 5(b): computed influence line for L 1 –U 2 , plotted with compression upward as on the examination paper, together with the idealised vehicle. See the official exam paper.]

Question 5 — influence-line results
Quantity	Ordinates	Largest absolute value
(a)(i) $M$ just right of B	0 at A and B, −2.00 m at C, 0 at D	2.00 m (hogging)
(a)(ii) $V$ just right of B	0 on A–B, +1.00 from B⁺ to C, 0 at D	1.00
(a)(iii) $H$ at E	−0.50 at A, 0 at B, +1.00 at C, 0 at D	1.00
(b) IL for L₁–U₂	0, +0.650 T, −1.300 C, −0.650 C, 0 at L₁ to L₅	1.300
(b)(i) Maximum compression	Lead 100 kN axle at L₃	260.0 kN
(b)(ii) Maximum tension	Second 100 kN axle at L₂	43.3 kN

Topic. Influence lines for determinate frames and trusses, and the placement of a moving load train for maximum effect.

Key relations. An influence line for a determinate structure is piecewise linear, so it is fully described by its panel-point ordinates. For a truss diagonal cut by a vertical section, $F = \pm (L_d/h)\,V_{\text{panel}}$. The effect of a load train is $\sum P_i y_i$, and because that sum is piecewise linear in the train position its extreme always occurs with some axle standing on a vertex.

Why this works. An influence line is the graph of one response quantity against the position of a unit load, so superposition lets any set of loads be handled by simple multiplication and addition. For a truss the load reaches the structure only at panel points, which is why the line is straight between them; for a determinate frame every reaction is a linear function of the load position, so the same linearity follows. The vertex rule turns what looks like a continuous search into a short finite list of candidate positions — here only ten — and that is what makes the question tractable in the marks allowed.

Common pitfall. Assuming that the heaviest axle belongs on the peak. In part (b) the tension case is governed by the second axle sitting on the tensile peak, because the trailing 50 kN then lands on the compressive peak and the leading axle picks up useful tension — a placement that gives 43.3 kN against only 10.8 kN for the obvious arrangement. A second trap in part (a) is to forget the unit jump in the shear influence line at the section.

Source. Hibbeler, Structural Analysis, Ch. 6 (beams and frames) and §6.6–6.8 (trusses and moving loads); Kassimali, Structural Analysis, Ch. 8–9.

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